2018 AMC 10A Problem 14

Attempt Problem 14 of the 2018 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10A solutions, or check the answer key.

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14.

What is the greatest integer less than or equal to 3100+2100396+296?\dfrac{3^{100}+2^{100}}{3^{96}+2^{96}}?

8080

8181

9696

9797

625625

Answer: A
Concepts:exponentfloor and ceiling functionsbounding to limit cases
Difficulty rating: 1540
Solution:

Let a=396a=3^{96} and b=296b=2^{96}. The expression is 81a+16ba+b=16+65aa+b\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}, so it is less than 16+65=8116+65=81.

To show the floor is 8080, we also need the expression to be greater than 8080. This is equivalent to 81a+16b>80a+80b81a+16b>80a+80b, or a>64ba>64b.

Because (32)2=94>2,\left(\dfrac32\right)^2=\dfrac94>2, we have ab=(32)96>248>64.\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64. Hence the expression is greater than 8080 and less than 81.81. Thus, A is the correct answer.

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